ABC is same as FDE
so AC = FE = 5ft
angle B = 28 deg
angle C = 90 deg
so A = 180 - B - C = 62
<u>Complete Question:</u>
Janeel has a 10 inch by 12 inch photograph. She wants to scan the photograph, then reduce the results by the same amount in each dimension to post on her Web site. Janeel wants the area of the image to be one eight of the original photograph. Write an equation to represent the area of the reduced image. Find the dimensions of the reduced image.
<u>Correct Answer:</u>
A) 
B) Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch
<u>Step-by-step explanation:</u>
a. Write an equation to represent the area of the reduced image.
Let the reduced dimensions is by x , So the new dimensions are

According to question , Area of new image is :
⇒ 
⇒ 
⇒ 
So the equation will be :
⇒ 
b. Find the dimensions of the reduced image
Let's solve : 
⇒ 
⇒ 
⇒ 
By Quadratic formula :
⇒ 
⇒ 
⇒ 
⇒
x = 15 is rejected ! as 15 > 10 ! Side can't be negative
⇒ 
Therefore, Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch
Answer:
15 seconds
Step-by-step explanation:
If you make a table of values for the dog and the squirrel using d = rt, then the rates are easy: the dog's rate is 150 and the squirrel's is 100. The t is what we are looking for, so that's our unknown, and the distance is a bit tricky, but let's look at what we know: the dog is 200 feet behind the squirrel, so when the dog catches up to the squirrel, he has run some distance d plus the 200 feet to catch up. Since we don't know what d is, we will just call it d! Now it seems as though we have 2 unknowns which is a problem. However, if we solve both equations (the one for the dog and the one for the squirrel) for t, we can set them equal to each other. Here's the dog's equation:
d = rt
d+200 = 150t
And the squirrel's:
d = 100t
If we solve both for t and set them equal to each other we have:

Now we can cross multiply to solve for d:
150d = 100d + 20,000 and
50d = 20,000
d = 400
But we're not looking for the distance the squirrel traveled before the dog caught it, we are looking for how long it took. So sub that d value back into one of the equations we have solved for t and do the math:

That's 1/4 of a minute which is 15 seconds.